Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

Biological case studies are a key part of the WJEC Year 12 examination, testing your ability to interpret data, apply knowledge to unfamiliar contexts, and evaluate experimental procedures. This article provides a series of practical exercises that will help you master case study questions.

生物案例研究是WJEC 12年级考试的重要组成部分，考查你解读数据、将知识应用于陌生情境以及评估实验步骤的能力。本文提供一系列实战练习，帮助你掌握案例分析题型。

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1. Core Skills for Case Study Analysis | 案例分析核心技能

Before diving into specific examples, it is essential to master the fundamental techniques. Always begin by carefully reading the stem of the question, noting the independent variable, dependent variable, and any control variables mentioned. Look for units and the range of data provided. When describing a graph or table, start with the overall trend, then quote specific data points to support your description. Use comparative language such as ‘increase’, ‘decrease’, ‘plateau’ and ‘optimum’. Finally, when evaluating a method, consider reliability, accuracy, and validity.

在深入具体案例之前，掌握基本技巧至关重要。首先仔细阅读题干，标注自变量、因变量和任何提到的控制变量。注意单位和所提供数据的范围。在描述图表或表格时，先概述整体趋势，再引用具体数据点来支持你的描述。使用比较性语言，如“上升”、“下降”、“趋于平稳”和“最适”。最后，在评估方法时，要考虑可靠性、准确性和有效性。

Another key skill is calculating percentage change and rates. Percentage change = ((final value – initial value) / initial value) × 100. Rate is often calculated as change in quantity divided by time. When plotting graphs, remember to label axes with quantity and unit, use an appropriate scale, and draw a line of best fit where relevant.

另一项关键技能是计算百分比变化和速率。百分比变化 = ((最终值 – 初始值) / 初始值) × 100。速率通常以变化量除以时间来计算。绘制图表时，记得标明坐标轴及其单位和物理量，使用合适的刻度，并在必要时画出最佳拟合线。

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2. Case Study 1: Temperature and Enzyme Activity | 案例一：温度与酶活性

An experiment investigated the effect of temperature on the activity of the enzyme amylase. Starch was added to an amylase solution at various temperatures, and the rate of product formation (maltose) was measured in mg min⁻¹. The results are shown in the table below.

一个实验探究了温度对淀粉酶活性的影响。在不同温度下向淀粉酶溶液中加入淀粉，并测量产物（麦芽糖）的生成速率，单位为 mg min⁻¹。结果如下表所示。

Temperature (°C)	Rate (mg min⁻¹)
0	0.5
10	1.2
20	2.8
30	5.4
40	7.0
50	3.2
60	0.1

Task: Plot a graph of rate against temperature. Describe and explain the shape of the graph. Calculate the percentage decrease in rate from the optimum temperature to 60 °C.

任务：绘制速率-温度曲线图。描述并解释曲线的形状。计算从最适温度到60 °C时速率的百分比下降。

Answer guidance: The graph should show a bell-shaped curve. The rate increases from 0 °C to 40 °C because the enzyme and substrate molecules have more kinetic energy, making successful collisions more frequent. 40 °C is the optimum temperature. Beyond this, the rate drops sharply because the enzyme denatures – the tertiary structure is disrupted by breaking hydrogen and ionic bonds, altering the active site so the substrate no longer fits. Percentage decrease = ((7.0 – 0.1) / 7.0) × 100 = 98.6%.

答题指导：曲线应呈钟形。速率从0 °C到40 °C上升，因为酶和底物分子的动能增加，成功碰撞更频繁。40 °C为最适温度。超过该温度，速率急剧下降，因为酶变性——氢键和离子键断裂，三级结构被破坏，活性位点改变，底物不再匹配。百分比下降 = ((7.0 – 0.1) / 7.0) × 100 = 98.6%。

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3. Case Study 2: Osmosis and Potato Cylinders | 案例二：渗透作用与土豆条

Potato cylinders of equal length and mass were placed in sucrose solutions of different concentrations for 24 hours. Their mass was measured before and after, and the percentage change in mass calculated. Any negative value indicates a net loss of water from the tissue.

将长度和质量相等的土豆条放入不同浓度的蔗糖溶液中24小时。测量其前后的质量，并计算质量变化百分比。负值表示组织净失水。

Sucrose concentration (mol dm⁻³)	Percentage change in mass (%)
0.0	+15.2
0.2	+8.4
0.4	+1.8
0.6	-4.6
0.8	-9.9
1.0	-14.1

Questions: Plot the data and use the graph to estimate the water potential of the potato tissue. Explain why the mass increased in pure water and decreased in 1.0 mol dm⁻³ sucrose. Suggest two improvements to increase the validity of the results.

问题：绘制数据曲线，并利用图表估算土豆组织的水势。解释为什么在纯水中质量增加，而在1.0 mol dm⁻³蔗糖溶液中质量减少。提出两项可提高结果有效性的改进措施。

Solution outline: The graph is a line crossing the x-axis where percentage change is zero; this corresponds to approximately 0.48 mol dm⁻³ sucrose, which is isotonic to the cell sap. The water potential of potato is equal to that of 0.48 mol dm⁻³ sucrose solution at atmospheric pressure. In pure water, the external water potential is higher than inside the cells, so water enters by osmosis, increasing mass. In concentrated sucrose, water leaves the cells, causing a decrease in mass. Improvements: blot the cylinders dry before weighing to remove surface water; use a larger number of replicates for each concentration to calculate a mean.

解答要点：曲线是一条穿过x轴的直线，在质量变化为零时对应大约0.48 mol dm⁻³蔗糖，此时溶液与细胞液等渗。土豆的水势与该浓度蔗糖溶液在大气压下的水势相等。在纯水中，外界水势高于细胞内部，水通过渗透作用进入细胞，质量增加。在浓蔗糖溶液中，水分离开细胞，导致质量下降。改进措施：称重前用滤纸吸干土豆条表面的水分；每个浓度使用多个重复样品并计算平均值。

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4. Case Study 3: Photosynthesis and Light Intensity | 案例三：光合作用与光照强度

An aquatic plant was placed in a beaker of water, and the number of oxygen bubbles released per minute was counted at different distances from a lamp. The lamp’s brightness was kept constant. The results are shown below.

将一株水生植物放在盛水的烧杯中，在距离灯不同位置处计数每分钟释放的氧气泡数目。灯的亮度保持不变。结果如下。

Distance from lamp (cm)	Bubbles per minute
10	45
20	28
30	18
40	18
50	17

Task: Explain why the bubble count drops with increasing distance. Identify the limiting factor at distances greater than 30 cm. Suggest why the bubble count does not fall to zero at 50 cm. Describe how you could modify the experiment to investigate the effect of carbon dioxide concentration.

任务：解释为什么气泡数随距离增加而减少。指出在距离大于30 cm时的限制因子。解释为何在50 cm处气泡数并未降至零。描述如何改进实验以探究二氧化碳浓度的影响。

Key answers: Light intensity decreases with distance because the same light energy spreads over a larger area (inverse square law). Thus, less ATP and NADPH are produced in the light-dependent reaction, so the Calvin cycle slows, reducing oxygen release. At distances beyond 30 cm, the bubble count remains almost constant, indicating that light intensity is no longer the limiting factor; carbon dioxide concentration or temperature may now be limiting. The bubbles do not stop completely because respiration continues in the plant, producing some carbon dioxide which can be used in photosynthesis. To test CO₂ effect, add sodium hydrogencarbonate solution at different concentrations to the water while keeping light intensity constant.

关键解答：光照强度随距离增加而降低，因为同样的光能散布到更大面积上（平方反比定律）。因此，光反应产生的ATP和NADPH减少，卡尔文循环减慢，氧气释放减少。在距离超过30 cm时，气泡数几乎恒定，表明光照强度不再是限制因子；此时二氧化碳浓度或温度可能成为限制因子。气泡并未完全停止，因为植物呼吸作用仍在进行，产生二氧化碳可用于光合作用。要探究CO₂的影响，可向水中加入不同浓度的碳酸氢钠溶液，同时保持光照强度恒定。

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5. Case Study 4: Pedigree Analysis of a Genetic Disorder | 案例四：遗传病家系图分析

The following pedigree shows the inheritance of a rare genetic disorder in a family. Filled symbols represent affected individuals. (A text-based description: In generation I, the father is affected, mother unaffected. They have three children: two daughters, both unaffected, and one son, affected. The son marries an unaffected woman, and they have two unaffected sons and one affected daughter.)

以下系谱展示了一种罕见遗传病在一个家族中的遗传情况。实心符号代表患病个体。（文字描述：第I代，父亲患病，母亲未患病。他们有三个孩子：两个女儿均未患病，一个儿子患病。患病儿子与一未患病女性结婚，生育两个未患病儿子和一个患病女儿。）

Questions: Determine whether the disorder is dominant or recessive. Is it autosomal or X-linked? Justify your answer. What is the probability that the affected son’s next child will be affected if the mother is a carrier?

问题：判断该疾病是显性还是隐性。是常染色体遗传还是X连锁遗传？论证你的答案。如果该患病儿子的妻子是携带者，他们下一个孩子患病的概率是多少？

Reasoning: The disorder is recessive because affected individuals appear in children of unaffected parents (the affected son in generation II has an affected father but unaffected mother, and skip-generation pattern). It is autosomal recessive, not X-linked, because an affected father (gen I) passed the trait to his son – if X-linked recessive, the father would pass his X chromosome only to daughters, not to a son. The affected son (II-3) must be homozygous recessive (aa). His wife is stated to be a carrier, so her genotype is Aa. Their next child has a 50% chance of being affected (aa) – the Punnett square shows ½ probability.

推理：该疾病为隐性遗传，因为患病个体出现在未患病的父母所生的孩子中（第II代患病儿子的父亲患病，但母亲未患病，且存在隔代遗传）。它是常染色体隐性而非X连锁隐性，因为患病父亲（I代）将性状传递给儿子——如果是X连锁隐性，父亲只会将X染色体传给女儿。患病儿子（II-3）必定是隐性纯合子（aa）。其妻子被指定为携带者，基因型为Aa。他们下一个孩子患病的概率为50%（aa）——旁氏表显示1/2概率。

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6. Case Study 5: Population Growth of Yeast | 案例五：酵母种群增长

A yeast culture was grown in a flask with a nutrient medium at 25 °C. Samples were taken every 2 hours, and the number of cells per mm³ was counted using a haemocytometer. The data are given below.

在25 °C下，酵母培养物在含有营养液的锥形瓶中生长。每2小时取样，用血球计数板计数每mm³的细胞数。数据如下。

Time (hours)	Cells per mm³ (×10³)
0	0.5
2	0.5
4	1.0
6	2.5
8	6.2
10	12.0
12	12.1
14	12.0

Tasks: Plot a growth curve. Label the lag, exponential (log) and stationary phases. Explain the biological reasons for each phase. Predict what would happen if the culture was continued for another 10 hours without adding fresh medium.

任务：绘制生长曲线。标出延滞期、指数（对数）期和稳定期。解释各阶段的生物学原因。预测如果不添加新鲜培养基继续培养10小时会发生什么。

Explanation: Lag phase (0–2 h): yeast cells adjust to the environment, synthesising enzymes and preparing for division. Exponential phase (4–10 h): cells divide rapidly by mitosis, with plenty of nutrients and space, no limiting factors. Stationary phase (12 h+): population levels off because nutrients become depleted, waste products (ethanol) accumulate, and space becomes limiting; birth rate equals death rate. If incubation continued, a death phase would occur: cells would die faster than they are produced, and the curve would decline. Some cells may survive using autophagy.

解释：延滞期（0–2小时）：酵母细胞适应环境，合成酶并准备分裂。指数期（4–10小时）：细胞通过有丝分裂迅速繁殖，营养物质和空间充足，无限制因素。稳定期（12小时以上）：群体数量趋于平稳，因为营养物质耗尽，代谢废物（乙醇）积累，空间受限；出生率等于死亡率。如果继续培养，将进入衰亡期：细胞死亡速率超过新生速率，曲线下降。部分细胞可能通过自噬存活。

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7. Case Study 6: Antibiotic Resistance in Bacteria | 案例六：细菌抗生素耐药性

A student investigated the effect of the antibiotic tetracycline on two strains of bacteria, A and B. Sterile discs soaked in the same concentration of tetracycline were placed on agar plates seeded with each strain. After incubation, the clear zones (zones of inhibition) were measured.

一名学生研究了四环素对抗生素对A、B两种菌株的影响。浸泡了相同浓度四环素的灭菌纸片放置在涂布了各菌株的琼脂平板上。培养后，测量透明圈（抑菌圈）直径。

Strain	Mean zone of inhibition (mm)
A	24.5
B	8.3

Questions: Which strain is more resistant to tetracycline? Explain the genetic basis for the resistance in strain B. Describe how natural selection could lead to an increase in the proportion of resistant bacteria in a population exposed to antibiotics. Suggest why antibiotics should be taken for the full prescribed course.

问题：哪种菌株对四环素更具耐药性？解释菌株B耐药性的遗传基础。描述在接触抗生素的群体中，自然选择如何导致耐药菌比例增加。说明为何应遵医嘱全程服用抗生素。

Answers: Strain B is more resistant because it has a much smaller zone of inhibition. Resistance arises from a mutation in the bacterial DNA, or via horizontal gene transfer (conjugation) receiving a plasmid carrying a resistance gene. This gene may code for an enzyme that breaks down the antibiotic or a protein that pumps the antibiotic out. When tetracycline is applied, susceptible bacteria die, but resistant ones survive and reproduce, passing the resistance allele to offspring. Over many generations, the population shifts to mostly resistant individuals. Completing the full course ensures all bacteria are killed, reducing the chance that partially resistant survivors multiply and develop full resistance.

解答：菌株B更耐药，因为其抑菌圈直径小得多。耐药性来源于细菌DNA的突变，或通过水平基因转移（接合）接收了携带耐药基因的质粒。该基因可能编码分解抗生素的酶，或可将抗生素泵出的蛋白。施用四环素时，敏感菌死亡，而耐药菌存活并繁殖，将耐药等位基因传给后代。经过多代，群体中大部分变为耐药个体。全程服药可确保所有细菌被杀死，降低部分耐药存活者增殖并发展为完全耐药的风险。

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8. Common Mistakes and Examiner Tips | 常见错误与考官提示

Many students lose marks by failing to use data from the table or graph in their explanations. Always quote figures (e.g., ‘the rate increased from 2.8 to 7.0 mg min⁻¹’) rather than giving a vague statement. Another common error is confusing ‘describe’ with ‘explain’. Description requires patterns and trends; explanation demands biological mechanisms. When a question asks you to ‘evaluate’, you must give both advantages and limitations of a method or conclusion. Additionally, always check units and significant figures – for example, calculating a rate as 0.035 cm min⁻¹, not 0.035 cm/min.

许多学生因未能在解释中引用表格或图中的数据而失分。务必引用具体数字（例如，“速率从2.8 mg min⁻¹ 升至 7.0 mg min⁻¹”），而非笼统表述。另一个常见错误是混淆“描述”与“解释”。描述要求概括模式和趋势；解释则要求阐明生物学机制。当题目要求“评价”时，必须给出方法或结论的优点和局限性。此外，务必检查单位和有效数字——例如，速率应表示为0.035 cm min⁻¹，而非0.035 cm/min。

In genetics cases, clearly define symbols and show a Punnett square or probability calculation. For data-based questions, pay attention to anomalies and suggest how to deal with them (repeat measurement, calculate mean excluding anomaly). Also, when discussing validity, mention controlling variables such as temperature, pH, and volume of solutions. Finally, time management is critical: allocate roughly 1.5 minutes per mark, and do not spend too long on graph plotting.

在遗传案例中，清楚定义符号并展示旁氏表或概率计算。对于基于数据的题目，留意异常值并提出处理方法（重复测量，排除异常值后计算平均值）。此外，在讨论有效性时，要提及控制变量，如温度、pH和溶液体积。最后，时间管理至关重要：大致按每1分1.5分钟分配时间，不要在图表的绘制上花费过长时间。

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9. Self-Assessment Practice Question | 自评练习题

Try this practice case study: A student measured the heart rate of a volunteer before and after a step-up exercise performed for 5 minutes. The resting heart rate was 72 bpm. It rose to 125 bpm immediately after exercise, fell to 90 bpm after 1 minute of rest, and returned to 72 bpm after 5 minutes. Another volunteer with higher fitness levels had a resting heart rate of 58 bpm, rose to 95 bpm after the same exercise, and recovered in 2 minutes. Explain the physiological reasons for the changes in heart rate during exercise and recovery. Compare the fitness levels of the two volunteers using the data. Suggest how the student could improve the reliability of the experiment.

试做以下案例研究：一名学生测量了一名志愿者在台阶运动前和运动5分钟后的心率。静息心率为72 bpm。运动后立即升至125 bpm，休息1分钟后降至90 bpm，5分钟后恢复至72 bpm。另一名体能水平较高的志愿者静息心率为58 bpm，同样运动后升至95 bpm，并在2分钟内恢复。解释运动期间和恢复期心率变化的生理原因。利用数据比较两名志愿者的体能水平。提出学生可如何提高实验的可靠性。

Model answers: During exercise, muscles contract more frequently, requiring more ATP. This increases aerobic respiration, which demands more oxygen and produces more carbon dioxide. Chemoreceptors detect the rise in CO₂, and the medulla oblongata sends impulses via the sympathetic nerve to increase heart rate (and stroke volume). After exercise, heart rate remains elevated to repay the oxygen debt, removing lactate and replenishing ATP/creatine phosphate. Volunteer 2 has a lower resting heart rate and a smaller increase during exercise, indicating a higher stroke volume; recovery is faster, showing better cardiovascular fitness. To improve reliability, the student should use multiple volunteers with similar age and gender, repeat the trial several times, and control step height and pace.

参考答案：运动时肌肉收缩更频繁，需要更多ATP。这增加了有氧呼吸，需氧量上升并产生更多二氧化碳。化学感受器检测到CO₂升高，延髓通过交感神经发出冲动，提高心率（及每搏输出量）。运动后心率仍较高，以偿还氧债，清除乳酸并补充ATP/磷酸肌酸。志愿者2静息心率更低，运动时增加幅度更小，表明每搏输出量更大；恢复更快，显示心血管功能更佳。为提高可靠性，学生应使用多名年龄和性别相似的志愿者，重复试验多次，并控制台阶高度和节拍。

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10. Conclusion and Revision Strategy | 总结与复习策略

Mastering case study questions requires regular practice with a variety of data formats: tables, line graphs, bar charts and diagrams. Each time you complete a past paper question, focus on the mark scheme language – note how examiners want you to phrase descriptions and explanations. Build a personal glossary of command words (describe, explain, suggest, evaluate, calculate) and their requirements. Create your own mini-case studies using data from practical work or textbook examples, and swap with friends for peer marking. With consistent effort, you will develop the analytical and evaluative skills that the WJEC exam demands.

要掌握案例分析题，需要定期练习各种数据形式：表格、折线图、条形图和示意图。每次完成一道历年真题后，关注评分细则的语言——注意考官希望如何表述描述与解释。建立个人指令词词汇表（描述、解释、建议、评价、计算）及其要求。利用实验数据或教材示例自拟小型案例研究，与同学交换批改。通过持续努力，你将发展WJEC考试所要求的分析和评价能力。

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